For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: increasing and decreasing functions, stationary points including stationary inflexion, second-derivative classification, tangents, normals, connected rates, and optimisation form the main route.
School-sensitive extension: unfamiliar related-rate geometry and multi-constraint optimisation may vary by school.
2027 national comparison: K341 Topic C1 includes monotonicity, stationary points, the second-derivative test, gradients, tangents, normals, connected rates, maxima, and minima.
Check your school: confirm sign-table, second-derivative, and justification conventions.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 16) Applications of Differentiation cover? A: Tangents, normals, stationary points, and related rates in IP AMaths.
Apply derivatives to describe gradient, optimise functions, and connect rates.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Applications of differentiation turn gradients into decisions.
Use it as a working check: Use derivatives to find tangent gradients, normal gradients, stationary points, and related rates. Always classify stationary points before naming a maximum or minimum.
Then go one layer deeper: Example: if y' = 0 at x = 3, test the second derivative or a sign chart before calling it a turning point. If y'' is positive, it is a local minimum.
Choosing the application route
Start by deciding what the derivative represents in the question. A derivative can be a gradient, a condition for a stationary point, or a link between changing quantities.
Question cue
First derivative step
What must be checked
Tangent at x=a
Find y′, then substitute x=a for the tangent gradient.
Use the curve to find the point of contact before writing the line.
Normal at x=a
Find the tangent gradient first.
Handle horizontal tangents and vertical normals separately.
Maximum or minimum value
Set y′=0, solve for candidates, then classify.
Check the required domain and use y′′ or a sign chart before naming the point.
Related rates
Write an equation linking the quantities, then differentiate with respect to time.
Convert the final answer to the rate the question actually asks for.
Common trap:y′=0 only says the curve is stationary. It does not by itself prove a maximum, minimum, or point of inflexion.
1 Common tasks
Tangent gradient at x=a: evaluate y′ at a.
Normal gradient: mnormal=−mtangent1 for non-zero tangents.
Stationary points satisfy y′=0; use second derivative or sign chart for classification.
Related rates: link quantities via differentiation with respect to time.
Tangent and normal line checkpoint
For tangent and normal questions, do not stop at the gradient. A line answer needs both the gradient and the point of contact.
Step
Tangent line
Normal line
Common trap
1. Differentiate
Find y′.
Find y′ first, because the normal depends on the tangent gradient.
Starting with the reciprocal before finding the tangent gradient.
2. Substitute the contact value
Put x=a into y′ to get mtangent
3. Find the point
Put x=a into the original curve to get (a,y(a)).
The normal also passes through (a,y(a)).
4. Write the line
Use y−y1=m(x−x1).
Use mnormal=−1/mtangent
Worked check: for y=x2+1 at x=2, y′=2x, so the tangent gradient is 4 and the point is (2,5). The tangent is y−5=4(x−2). The normal gradient is −41, so the normal is y−5=−41(x−2).
Misconception check: if mtangent=0, the tangent is horizontal and the normal is a vertical line x=a, not a line with gradient −1/0.
Related-rates setup checkpoint
For related-rates questions, write the quantity relationship before differentiating. The derivative step is usually a chain-rule step, not a new formula to memorise.
Setup question
What to write
Common trap
Which quantities are linked geometrically?
An equation such as area, volume, or Pythagoras that connects them.
Differentiating before removing unnecessary variables.
Which rate is given?
Mark it as a derivative with respect to time, such as dtdV.
Treating dtdV as if it were just V.
Which rate is required?
Rearrange after differentiating to isolate the requested derivative.
Giving dtdV when the question asks for dtdh
At what instant is the rate needed?
Substitute the instant-specific length, radius, height, or volume after differentiating.
Substituting changing values too early and turning variables into constants.
Worked check: if a sphere has V=34πr3, then dtdV=4πr2dtdr. If the volume is increasing at 12cm3⋅s−1 when r=3cm, then 12=4π(3)2dtdr, so dtdr=3π1,cm⋅s−1.
Misconception check: a negative rate means the quantity is decreasing in time. It does not mean the length, area, or volume itself is negative.
2 Worked example - Tangent and normal
For y=x3−3x+2 find the tangent and normal at x=1.
Derivative: y′=3x2−3.
Gradient at x=1: mt=0 (horizontal tangent).
Point: y=1−3+2=0 so coordinate (1,0).
Tangent: y=0.
Normal would require reciprocal gradient but mt=0⟹ normal is vertical line x=1.
3 Worked example - Optimisation
Find the minimum value of y=x+x9 for x>0.
Derivative: y′=1−x29.
Set y′=0: 1=x29 → x2=9 → x=3 (positive regime).
Second derivative: y′′=x318>0 for x>0
Minimum value: y=3+39=6.
Stationary-point classification checkpoint
After solving y′=0, classify the point before writing the final description. The classification step decides whether the answer is a maximum, minimum, or stationary point of inflexion.
Evidence
Classification
What to write
y′ changes from positive to negative.
Local maximum.
The curve rises before the point and falls after it.
y′ changes from negative to positive.
Local minimum.
The curve falls before the point and rises after it.
y′ keeps the same sign on both sides.
Stationary point of inflexion.
The curve flattens but does not turn.
y′′>0 at the stationary point.
Local minimum.
The curve is concave upward there.
y′′<0 at the stationary point.
Local maximum.
The curve is concave downward there.
y′′=0 at the stationary point.
Test is inconclusive.
Use a sign chart for y′ instead of guessing.
Worked check: for y=(x−1)3, y′=3(x−1)2, so y′=0 at x=1. But y′ is positive on both sides of x=1, so the curve keeps increasing. The point is a stationary point of inflexion, not a maximum or minimum.
Common trap:y′′=0 does not prove a point of inflexion by itself. It only means the second-derivative test has not decided the classification.
